This particular equation is a special case of Mihailescu's theorem: rewriting it as $2^x - 3^y = 1$, we know that either $x \le 1$ or $y \le 1$, because the only instance of higher powers having difference $1$ is $3^2 - 2^3$. This gives us the solutions $(x,y) = (2,1)$ and $(x,y) = (1,0)$.
In general, exponential Diophantine equations like this one can be solved by tedious application of the technique "take the equation modulo some arbitrary number like $37$ or something".
It's not quite that bad, though: we would just look for values of $m$ such that $2$ or $3$ have small multiplicative order mod $m$, i.e., $m$ divides $2^k-1$ or $3^k-1$ for some small $k$. This would either tell us that no solutions exist or give us some modular conditions on $x$ and $y$ which form part of a contradiction.
If some small solutions do exist, we would have to consider the equation modulo $2^k$ or $3^k$ to eliminate larger solutions. For example, in this problem, we have solutions for $x=1$ and $x=2$. If we had a solution in which $x \ge 3$, we would have $2^x \equiv 0 \pmod 8$. In that case, $$3^y \equiv 2^x - 1 \equiv 7 \pmod 8$$ but this is impossible: $3^y \bmod 8$ is $3$ when $y$ is odd, and $1$ when $y$ is even. So there can be no solution with $x \ge 3$.
(It's easy to check $x=0$, $x=1$, and $x=2$ to figure out that the above solutions are the only ones possible.)